\newproblem{lay:2_7_10}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.7.10}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Consider the following	geometric 2D transformations: $D$, a dilation (in which the $x$ and $y$ coordinates are scaled by the same factor); $R$, a rotation; and $T$,
	a translation. Does $D$ commute with $R$? That is ($D(R(\mathbf{x}))=R(D(\mathbf{x}))$ for all $\mathbf{x}\in\mathbb{R}^2$? Does $D$ commute with $T$? Does $T$ commute
	with $R$?
}{
  % Solution
	The three proposed transformations can be written as matrix transformations in homogeneous coordinates
	\begin{center}
		\begin{tabular}{l}
			$D(\mathbf{x})=\begin{pmatrix}r & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1\end{pmatrix}\tilde{\mathbf{x}}$ \\
			$R(\mathbf{x})=\begin{pmatrix}\cos(\alpha) & \sin(\alpha) & 0 \\ -\sin(\alpha) & \cos(\alpha) & 0  \\ 0 & 0 & 1\end{pmatrix}\tilde{\mathbf{x}}$ \\
			$T(\mathbf{x})=\begin{pmatrix}1 & 0 & \Delta x \\ 0 & 1 & \Delta y \\ 0 & 0 & 1\end{pmatrix}\tilde{\mathbf{x}}$ \\
		\end{tabular}
	\end{center}
	Now we need to check whether $D(R(\mathbf{x}))=R(D(\mathbf{x}))$
	\begin{center}
		$\begin{array}{rcl}D(R(\mathbf{x}))&=&
		   D\left(\begin{pmatrix}\cos(\alpha) & \sin(\alpha) & 0 \\ -\sin(\alpha) & \cos(\alpha) & 0  \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}\right)\\
		   &=&D\left(\begin{pmatrix}\cos(\alpha)x+\sin(\alpha)y\\-\sin(\alpha)x+\cos(\alpha)y\\1\end{pmatrix}\right)
			 =\begin{pmatrix}r & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}\cos(\alpha)x+\sin(\alpha)y\\-\sin(\alpha)x+\cos(\alpha)y\\1\end{pmatrix}\\
			 &=&\begin{pmatrix}r\cos(\alpha)x+r\sin(\alpha)y\\-r\sin(\alpha)x+r\cos(\alpha)y\\1\end{pmatrix}
		\end{array}$
	\end{center}
	On the other side
	\begin{center}
		$\begin{array}{rcl}R(D(\mathbf{x}))&=&
		   R\left(\begin{pmatrix}r & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}\right)\\
		   &=&R\left(\begin{pmatrix}rx\\ry\\1\end{pmatrix}\right)
			 =\begin{pmatrix}\cos(\alpha) & \sin(\alpha) & 0 \\ -\sin(\alpha) & \cos(\alpha) & 0  \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}rx\\ry\\1\end{pmatrix}\\
			 &=&\begin{pmatrix}r\cos(\alpha)x+r\sin(\alpha)y\\-r\sin(\alpha)x+r\cos(\alpha)y\\1\end{pmatrix}
		\end{array}$
	\end{center}
	So $D(R(\mathbf{x}))=R(D(\mathbf{x}))$ and rotation commutes with dilation.\\
	If we repeat the same exercise with dilations and translations
	\begin{center}
		$\begin{array}{rcl}D(T(\mathbf{x}))&=&
		   D\left(\begin{pmatrix}1 & 0 & \Delta x \\ 0 & 1 & \Delta y \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}\right)\\
		   &=&D\left(\begin{pmatrix}x+\Delta x\\y+\Delta y\\1\end{pmatrix}\right)
			 =\begin{pmatrix}r & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x+\Delta x\\y+\Delta y\\1\end{pmatrix}\\
			 &=&\begin{pmatrix}rx+r\Delta x\\ry+r\Delta y\\1\end{pmatrix}
		\end{array}$
	\end{center}
	On the other side
	\begin{center}
		$\begin{array}{rcl}T(D(\mathbf{x}))&=&
		   T\left(\begin{pmatrix}r & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}\right)\\
		   &=&T\left(\begin{pmatrix}rx\\ry\\1\end{pmatrix}\right)
			 =\begin{pmatrix}1 & 0 & \Delta x \\ 0 & 1 & \Delta y \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}rx\\ry\\1\end{pmatrix}\\
			 &=&\begin{pmatrix}rx+\Delta x\\ry+\Delta y\\1\end{pmatrix}
		\end{array}$
	\end{center}
	So $D(T(\mathbf{x}))\neq T(D(\mathbf{x}))$ and translation does not commute with dilation. Repeting once more the exercise with rotation and translation we would
	reach the conclusion that they do not commute.
}
\useproblem{lay:2_7_10}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
